If $a=\cos\frac{2π}7+i\sin\frac{2π}7$, $b=a+a^2+a^4$, $c=a^3+a^5+a^6$, show that $b$ and $c$ are the roots of the equation $x^2+x+2=0$.

  • $\begingroup$ Please see math.meta.stackexchange.com/questions/5020 for advice on how to effectively format your questions. $\endgroup$ – Lord Shark the Unknown Jan 6 '18 at 7:30
  • $\begingroup$ Though I can not read your question, I can see you have shown no effort so far. Try showing some effort so that we can help you figure out where you went wrong. $\endgroup$ – Mohammad Zuhair Khan Jan 6 '18 at 7:46
  • $\begingroup$ Now add what you have tried so far. $\endgroup$ – Mohammad Zuhair Khan Jan 6 '18 at 8:04
  • $\begingroup$ Hint: Substituting $a = e^{\frac{2 \iota \pi}7}$ might help. $\endgroup$ – Mohammad Zuhair Khan Jan 6 '18 at 8:08
  • $\begingroup$ take $e^{2i\pi/7}$ as a where a is 7th root of unity $\endgroup$ – Hydrous Caperilla Jan 6 '18 at 8:32


Note that



$$1+a+a^2+a^3+a^4+a^5+a^6=0$$ Proof that sum of complex unit roots is zero

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    $\begingroup$ +1 for providing the basic idea to the OP rather than solving it $\endgroup$ – user501655 Jan 6 '18 at 8:39
  • $\begingroup$ ops...sorry for the typo, thanks! $\endgroup$ – user Jan 6 '18 at 19:48

$$b+c=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\right)i+$$ $$+\cos\frac{6\pi}{7}+\cos\frac{10\pi}{7}+\cos\frac{12\pi}{7}+\left(\sin\frac{6\pi}{7}+\sin\frac{10\pi}{7}+\sin\frac{12\pi}{7}\right)i=$$ $$=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}+$$ $$+\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}-\sin\frac{6\pi}{7}+\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}\right)i=$$ $$=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$ $$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-1.$$ You can calculate the value of $ab$ by the same way.

Indeed, since we proved that $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2},$$ we obtain: $$bc=\left(-\frac{1}{2}+\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\right)i\right)\left(-\frac{1}{2}-\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\right)i\right)=$$ $$=\frac{1}{4}+\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}-\sin\frac{6\pi}{7}\right)^2=$$ $$=\tfrac{1}{4}+\sin^2\frac{2\pi}{7}+\sin^2\frac{4\pi}{7}+\sin^2\frac{6\pi}{7}+2\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}-2\sin\frac{2\pi}{7}\sin\frac{6\pi}{7}-2\sin\frac{4\pi}{7}\sin\frac{6\pi}{7}=$$ $$=\frac{1}{4}+\frac{1}{2}\left(1-\cos\frac{4\pi}{7}+1-\cos\frac{6\pi}{7}+1-\cos\frac{2\pi}{7}\right)+$$ $$+\cos\frac{2\pi}{7}-\cos\frac{6\pi}{7}+\cos\frac{6\pi}{7}-\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}-\cos\frac{2\pi}{7}=\frac{1}{4}+\frac{7}{4}=2$$ and use the Vieta's theorem.

Also, we can use De Moivre here.

$a^7=\cos2\pi+i\sin2\pi=1$ and since $a\neq1$, we obtain: $$a^6+a^5+a^4+a^3+a^2+a+1=0,$$ which gives $b+c=-1.$

Now, $$bc=a^4(1+a+a^3)(1+a^2+a^3)=a^4(a^6+a^5+a^4+3a^3+a^2+a+1)=2a^7=2$$ and we are done again.

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    $\begingroup$ nice way, original as usual! $\endgroup$ – user Jan 6 '18 at 8:37
  • $\begingroup$ Would you present this solution to your students after introducing the topic of roots of unity? $\endgroup$ – uniquesolution Jan 6 '18 at 11:04
  • $\begingroup$ @uniquesolution ago Yes of course. I added something. See now. $\endgroup$ – Michael Rozenberg Jan 6 '18 at 11:40
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Jan 7 '18 at 9:30

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